Difference between revisions of "2021 AMC 12A Problems/Problem 12"
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==Problem== | ==Problem== | ||
All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>? | All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>? | ||
Revision as of 18:41, 12 February 2021
- The following problem is from both the 2021 AMC 10A #14 and 2021 AMC 12A #12, so both problems redirect to this page.
Contents
Problem
All the roots of the polynomial 
are positive integers, possibly repeated. What is the value of 
?
Solution 1:
By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is 
. Therefore, 
.
~JHawk0224
Solution 2:
Using the same method as Solution 1, we find that the roots are 
and 
. Note that is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the 
products 
we obtain ![\[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A)} -88}.\]](http://latex.artofproblemsolving.com/c/9/4/c94501df13627524ea9bd8865283fed1b672f3e8.png)
~ ike.chen
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLern (Using Vieta's Formulas & Combinatorics)
~ pi_is_3.14
See also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
